Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 26416 Accepted: 13641

Description

A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:



In the figure, each node is labeled with an integer from {1, 2,…,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if node x is in the path between the root and node y. For example, node 4 is an ancestor of node 16. Node 10 is also an ancestor of node 16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7. A node x is called a common ancestor of two different nodes y and z if node x is an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is called the nearest common ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y and z among their common ancestors. Hence, the nearest common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is.

For other examples, the nearest common ancestor of nodes 2 and 3 is node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor of z, then the nearest common ancestor of y and z is y.

Write a program that finds the nearest common ancestor of two distinct nodes in a tree.

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,…, N. Each of the next N -1 lines contains a pair of integers that represent an edge –the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers whose nearest common ancestor is to be computed.

Output

Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.

Sample Input

2

16

1 14

8 5

10 16

5 9

4 6

8 4

4 10

1 13

6 15

10 11

6 7

10 2

16 3

8 1

16 12

16 7

5

2 3

3 4

3 1

1 5

3 5

Sample Output

4

3

Source

Taejon 2002

【题解】



裸的LCA。这道题用的是树上倍增;

即先让两个点到同一深度。然后再一起网上走到汇合点。(每次走2^j步);

记住这个递推公式

p[i][j] = p[p[i][j-1]][j-1];

#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm> using namespace std; const int MAXN = 20000;
const int MAX = 14; vector <int> son[MAXN];
int n,fa[MAXN],p[MAXN][MAX+5],dep[MAXN],pre[MAX+5]; void input(int &r)
{
char t = getchar();
while (!isdigit(t)) t = getchar();
r = 0;
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
} void dfs(int x,int f)
{
dep[x] = dep[f] + 1;
p[x][0] = f;
for (int i = 1; i <= MAX; i++)
p[x][i] = p[p[x][i - 1]][i - 1];
int len = son[x].size();
for (int i = 0; i <= len - 1; i++)
{
int y = son[x][i];
dfs(y, x);
}
} int main()
{
//freopen("F:\\rush.txt", "r", stdin);
pre[0] = 1;
for (int i = 1; i <= MAX; i++)
pre[i] = pre[i - 1] << 1;
int T;
input(T);
while (T--)
{
input(n);
for (int i = 1; i <= n; i++)
son[i].clear(),fa[i] = 0;
for (int i = 1; i <= n - 1; i++)
{
int x, y;
input(x); input(y);
son[x].push_back(y);
fa[y]++;
}
int root;
for (int i = 1; i <= n; i++)
if (fa[i] == 0)
{
root = i;
break;
}
dfs(root,0);
int t0, t1;
input(t0); input(t1);
if (dep[t0] > dep[t1])
swap(t0, t1);
for (int i = MAX; i >= 0; i--)
if (dep[t0] <= dep[t1] - pre[i])
t1 = p[t1][i];
if (t1 == t0)
{
printf("%d\n", t1);
continue;
}
for (int i = MAX; i >= 0; i--)
{
if (p[t0][i] == p[t1][i])
continue;
t0 = p[t0][i], t1 = p[t1][i];
}
printf("%d\n", p[t0][0]);
}
return 0;
}

【51.64%】【POJ 1330】Nearest Common Ancestors的更多相关文章

  1. POJ 1330:Nearest Common Ancestors

    Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20940   Accept ...

  2. POJ 1330 Nearest Common Ancestors 【LCA模板题】

    任意门:http://poj.org/problem?id=1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000 ...

  3. POJ - 1330 Nearest Common Ancestors(基础LCA)

    POJ - 1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000KB   64bit IO Format: %l ...

  4. POJ 1330 Nearest Common Ancestors / UVALive 2525 Nearest Common Ancestors (最近公共祖先LCA)

    POJ 1330 Nearest Common Ancestors / UVALive 2525 Nearest Common Ancestors (最近公共祖先LCA) Description A ...

  5. POJ.1330 Nearest Common Ancestors (LCA 倍增)

    POJ.1330 Nearest Common Ancestors (LCA 倍增) 题意分析 给出一棵树,树上有n个点(n-1)条边,n-1个父子的边的关系a-b.接下来给出xy,求出xy的lca节 ...

  6. POJ 1330 Nearest Common Ancestors (LCA,dfs+ST在线算法)

    Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14902   Accept ...

  7. LCA POJ 1330 Nearest Common Ancestors

    POJ 1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24209 ...

  8. POJ 1330 Nearest Common Ancestors(lca)

    POJ 1330 Nearest Common Ancestors A rooted tree is a well-known data structure in computer science a ...

  9. POJ 1330 Nearest Common Ancestors 倍增算法的LCA

    POJ 1330 Nearest Common Ancestors 题意:最近公共祖先的裸题 思路:LCA和ST我们已经很熟悉了,但是这里的f[i][j]却有相似却又不同的含义.f[i][j]表示i节 ...

  10. pku 1330 Nearest Common Ancestors LCA离线

    pku 1330 Nearest Common Ancestors 题目链接: http://poj.org/problem?id=1330 题目大意: 给定一棵树的边关系,注意是有向边,因为这个WA ...

随机推荐

  1. SPARK-SQL内置函数之字符串函数

    转载请注明转自:http://www.cnblogs.com/feiyumo/p/8763186.html 1.concat对于字符串进行拼接 concat(str1, str2, ..., strN ...

  2. [React Native]升级React Native版本

    React Native正式版本还没发布,但是小版本基本上每个月都更新1-2次.9月11号又更新了0.33版本,其中有两个增强功能正好是项目中用到的. 添加Android6.0权限验证API Add ...

  3. sql —— between

    BETWEEN 操作符在 WHERE 子句中使用,作用是选取介于两个值之间的数据范围. 原表: 执行查询: 上面就可以搜索出得分为80~90的学生了,包含80,也包含90.

  4. AtCoder Beginner Contest 079 D - Wall【Warshall Floyd algorithm】

    AtCoder Beginner Contest 079 D - Wall Warshall Floyd 最短路....先枚举 k #include<iostream> #include& ...

  5. this的作用

    1.在一般函数方法中使用 this 指代全局  function test(){ this.x = 1; alert(this.x);  }  test(); // 1 2.作为对象方法调用,this ...

  6. javascript中数字的一些常规操作

    1,禁止输入 - (减号.负号) // html <input type="number" class="no-negative"> // js $ ...

  7. Xcode编译报错:< Apple Mach-O Linker Warning > clang: error: no such file or directory: 'xxxx'

    Xcode编译报错概述: clang: error: no such file or directory: 'CoreGraphics' 一般原因是链接库内容导入丢失,这种的排查下target - B ...

  8. QT 建立信号和槽的联系(事件处理)

    Qt中事件处理机制叫做“信号”和“槽”signal &slot. 其模型为: 对象a中有一个信号signal:XXX(代表一个事件) 对象b中有一个槽slot:YYY(事件处理函数) 用con ...

  9. shell 解析json

    未完待续 ### 解析api json文件为csv文件 cd /api ` do id=$(echo ${i}|sed 's/.html//') echo -n "${id}|" ...

  10. 2019-2-24-VisualStudio-过滤输出窗口文本

    title author date CreateTime categories VisualStudio 过滤输出窗口文本 lindexi 2019-2-24 11:10:7 +0800 2019-0 ...